 Hello everyone. Welcome to the session of application of multiple integrals, part 5. This is Swati Nikam, assistant professor, department of humanities and sciences, Vaalchand Institute of Technology, Sulapur. At the end of this session, student will be able to find mass of lamina by using double integration. Before we proceed, students please pause your video and answer this question. Write the formula for finding mass of lamina in Cartesian coordinate system, which we have already studied in earlier video. So I hope you have written your answer. Now, mass of lamina in Cartesian coordinate system is if the surface density rho of plane varies from point to point of the lamina and if it can be expressed as a function of coordinates of a point, then the mass of elementary area dA is rho into dA. If rho is equal to f of xy and since dA is equal to dx into dy, so formula for mass of lamina in Cartesian coordinate system is given as m is equal to double integration over f of xy into dx into dy. Now friends, let's see mass of lamina in polar coordinate system. If the surface density rho of a plane varies from point to point of the lamina and if it can be expressed as a function of the coordinates of a point of mass of elementary area dA is rho into dA and if rho is equal to f of r theta since dA is equal to r dr d theta, so mass of lamina in polar coordinate system is capital M equal to double integration over f of r, theta into r dr d theta. Now let's solve an example. The density of a circular lamina x square plus y square is equal to 2x varies as the distance from the fixed point of the circumference of the circle find the mass of lamina. So in order to get this lamina, let's draw the given figure. We have to change Cartesian coordinates to polar form. So we will make a substitution x is equal to r cos theta and y is equal to r sin theta as well as dx into dy becomes r dr d theta. Now here the equation of circle x square plus y square is equal to 2x becomes in place of x it is r cos theta so x square is equal to r square cos square theta plus y square becomes now r square sin square theta is equal to 2x is equal to 2r cos theta. So taking r square common from left hand side it is r square into bracket cos square plus sin square theta is 1. Is equal to 2r cos theta which is again equal to cancelling r from both the sides r is equal to 2 cos theta. So the circle x square plus y square is equal to 2x in polar form becomes r is equal to 2 cos theta. We have drawn in this picture xy coordinate system. So the circle x square plus y square is equal to 2x now r is equal to 2 cos theta. This x axis is a initial line representing theta is equal to 0 and this y axis is a perpendicular line with theta is equal to pi by 2. Let us consider p as a fixed point from origin. Now let p of xy be any point on the lamina. It is given that the density varies as square of the distance of point p xy from a fixed point say origin. Now by using distance formula the distance op is given as square root of x minus 0 whole square plus y minus 0 whole square which is equal to under root of x square plus y square. Therefore density rho is equal to k into bracket under root of x square plus y square whole square as density varies with the square of the distance from the point to the origin. So this is square of the distance and k is proportionality constant which is equal to k into under root of x square plus y square. Here it is x square plus y square is simply r square. So r square bracket square is equal to square root and the square get cancelled and it is k into r square. So density is now k into r square. Now to find mass of lamina over circular region it is convenient to find it above the x axis and multiply it by 2. So we know that this circular region is symmetric above and below the initial line. So let us calculate the lamina above the region only and multiply it by 2. Therefore mass of lamina m is equal to double integration over r of rho of r theta into r dr d theta is equal to double integration over r into bracket. Substitute value of rho density we have calculated over here k into r square into r dr d theta. In this region r now r varies from let us know the limits of integration with the help of this region. So r in the region r varies from pole r is equal to 0 to boundary of the circle r is equal to 2 cos theta and theta varies from theta is equal to 0 to theta is equal to pi by 2. Therefore m equal to 2 times integration 0 to pi by 2 inner integration 0 to 2 cos theta k into r cube dr d theta. As the limits are in terms of theta so let us integrate with respect to r first which is equal to 2 into k is constant. So outside the integration sign integration of 0 to pi by 2 integration of r cube is r raise to 4 upon 4. Limits of this integration 0 to 2 cos theta into d theta which is equal to 2 into k into integration 0 to pi by 2 in place of r substitute upper limit 2 cos theta whole raise to 4 divided by 4 minus lower limit 0 outside the bracket d theta which is equal to 2k into integration 0 to pi by 2 bracket 2 cos theta whole raise to 4 upon 4 minus 0 outside the bracket d theta is equal to Now this is 2 raise to 4 let us have it outside multiplied by 2 is 2 raise to 5 into k divided by 4 into integration 0 to pi by 2 cos raise to 4 theta into d theta by using the formula of beta function this cos raise to 4 is reduced to the beta form as which is equal to 32 by 4 into k and now by using reduction formula for this integration 0 to pi by 2 cos raise to 4 theta d theta is equal to 1 by 2 times beta of 1 by 2 comma pi by 2 therefore m is equal to 32 by 4 into k into 1 by 2 beta of half comma pi by 2 now by using relation between beta and gamma function we can further reduce it to is equal to this 32 by 8 is 4 into k into bracket now let us use relation between beta and gamma beta of m n is equal to gamma of m into gamma of n divided by gamma of m plus n so it is gamma of half into gamma of 5 by 2 divided by gamma of half plus 5 by 2 which is equal to 4k into bracket gamma of half into gamma of 5 by 2 upon gamma of 3 so m is equal to 4k into bracket gamma of half into gamma of 5 by 2 into gamma of 3 is equal to 4k into we know that gamma of half is equal to root pi so root pi into gamma of 5 by 2 is 3 by 2 into 1 by 2 into gamma of half which is root pi whole divided by gamma of 3 is 2 factorial that is equal to this 4 into 3 is 12 root pi into root pi is pi into k divided by this 2 into 2 into 2 is 8 so mass of lamina m is equal to 3 by 2 into pi k I have referred textbook named advanced engineering mathematics by H. K. Das for creation of this video thank you so much for watching this video and have a happy learning